Câu 1: Trong phân tử este (X) no, đơn chức, mạch hở có thành phần oxi chiếm 36,36 % khối lượng. Số đồng phân bố

Determine the wavelength of the energy that needs to be absorbed for a 3p electron in chlorine to be promoted to the 4s subshell

bazaltina [42]

Answer:

The wavelength of the energy that needs to be absorbed = 52.36 nm

Explanation:

For this study;

Let consider the Rydgberg equation from Bohr's theory of atomic model:

i.e.

$\dfrac{1}{\lambda} = R_H (Z^*)^2( \dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})$

where

Z* = effective nuclear charge of atom = Z - σ = 6

n₁ = lower orbit = 3

n₂ = higher orbit = 4

$R_H$ = Rydyberg constant = 1.09 × 10⁷ m⁻¹

λ = wave length of the light absorbed

∴

$\dfrac{1}{\lambda} = 1.09 \times 10^7}(6)^2( \dfrac{1}{3^2}-\dfrac{1}{4^2})$

$\dfrac{1}{\lambda} = 1.09 \times 10^7}(36)( \dfrac{1}{9}-\dfrac{1}{16})$

$\dfrac{1}{\lambda} = 392400000\times0.0486111111$

$\dfrac{1}{\lambda} =19075000$

$\lambda = \dfrac{1}{19075000}$

$\lambda = \dfrac{1}{1.91\times 10^7 \ m^{-1}}$

$\lambda = 5.236 \times 10^{-8} m$

$\lambda = 52.36 \times 10^{-9} m$

$\lambda = 52.36\ n m$

Therefore, the wavelength of the energy that needs to be absorbed = 52.36 nm

7 0

3 years ago

A chromium oxide compound contains 104.0 grams of chromium and 48.0 grams of oxygen. What is the most likely empirical formula o

TEA [102]

Answer: Empirical formula of this compound is $Cr_2O_3$

Explanation:

Mass of Cr= 104.0 g

Mass of O = 48.0 g

Step 1 : convert given masses into moles.

Moles of Cr = $\frac{\text{ given mass of Cr}}{\text{ molar mass of Cr}}= \frac{104.0g}{52g/mole}=2moles$

Moles of O = $\frac{\text{ given mass of O}}{\text{ molar mass of O}}= \frac{48.0g}{16g/mole}=3moles$

Step 2 : For the mole ratio, divide each value of moles by the smallest number of moles calculated.

For Cr = $\frac{2}{2}=1$

For O = $\frac{3}{2}=1.5$

Converting into simple whole number ratios by multiplying by 2

The ratio of Cr : O= 2: 3

Hence the empirical formula is $Cr_2O_3$

The most likely empirical formula of this compound is $Cr_2O_3$

4 0

3 years ago

A catalyst -

cupoosta [38]

A catalyst increases the reactions rate

3 0

3 years ago

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what is the empirical formula for a compound that is 50.61% copper 11.16% nitrogen and 38.23% oxygen

jonny [76]

Answer:

CuNO3

Explanation:

In order to calculate the empirical formula in this question;

Cu = 50.61% = 50.61g

N = 11.16% = 11.16g

O = 38.23% = 38.23g

Next, we convert each gram unit to moles by dividing by their respective molar mass (Where; Cu = 63.55, N = 14, O = 16)

Cu = 50.61 ÷ 63.55 = 0.796mol

N = 11.16 ÷ 14 = 0.797mol

O = 38.23 ÷ 16 = 2.389mol

Next, we divide each mole value by the smallest (0.796)

Cu = 0.796mol ÷ 0.796 = 1

N = 0.797mol ÷ 0.796 = 1.001

O = 2.389mol ÷ 0.796 = 3.001

Approximately, the simple whole number ratio between Cu, N and O is 1:1:3, hence, the empirical formula is CuNO3.

8 0

3 years ago

How many compounds are in a wooden block

Ksju [112]

1.........................

5 0

3 years ago

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